I am interested in the following problem
Let $k\in \{1,2,\ldots,9\}$ and $N\in \mathbb N$. Let $a(N)$ be the number of integers $n=0,\ldots,N$ such that the number $2^n$ starts with the digit $k$ when written in the decimal system. Prove the following limit $$\lim_{N\rightarrow \infty}\frac{a(N)}{N+1} = \log(k+1)-\log(k)$$ where $\log$ refers to the decimal logarithm.
I believe that this result may be proved using the following application of Von Neumann's ergodic theorem, which I managed to prove
Let $E = L^2_{1}(\mathbb R)$ be the space of $1$-periodic mesurable functions, quotiented by the subspace of such functions which are almost everywhere zero. Let $\alpha \in \mathbb R\setminus \mathbb Q$ and $f \in E$. In $E$, we have the following convergence
$$f_N := \frac{1}{N+1}\sum_{k=0}^Nf(\,\cdot+k\alpha) \xrightarrow[N\to \infty]{} \hat{f}(0):=\int_0^1fd\lambda$$ where $\lambda$ is the Lebesgue measure and $\hat{f}(0)$ is the constant Fourier coefficient of $f$.
Let $I=[\log(k),\log(k+1)[$ and $f$ be the prolongation of the indicatrix function $\mathbf 1_I$ by periodicity over $\mathbb R$. Let $\alpha := \log(2)$. Note that a real number $x\geq 1$ has $k$ as its first digit in the decimal system if and only if $f(\log(x))=1$. Hence, we may write the following $$\frac{a(N)}{N+1} = \frac{1}{N+1}\sum_{k=0}^Nf(k\alpha)=f_N(0)$$
My problem is that I only know about the convergence of the sequence $f_N$ as elements of $L^2$, which does imply that there exists some subsequence which converges almost everywhere, but I have no information a priori on this set where it converges. Namely, I do not know whether $0$ belongs to this set or not.
I am looking for a proper way to justify that the sequence $f_N(0)$ converges in $\mathbb R$ to $\hat{f}(0) = \lambda(I) = \log(k+1)-\log(k)$. I thought that I could maybe make it work because the sequence $f_N$ converges in $L^2$ to a constant function, which reminded me of Slutsky's lemma in probability theory, but I was not able to conclude.
Any help would be greatly appreciated.